International Journal of Mathematics and Statistics Invention (IJMSI) is an international journal intended for professionals and researchers in all fields of computer science and electronics. IJMSI publishes research articles and reviews within the whole field Mathematics and Statistics, new teaching methods, assessment, validation and the impact of new technologies and it will continue to provide information on the latest trends and developments in this ever-expanding subject. The publications of papers are selected through double peer reviewed to ensure originality, relevance, and readability. The articles published in our journal can be accessed online.

1. International Journal of Mathematics and Statistics Invention (IJMSI)
E-ISSN: 2321 – 4767 P-ISSN: 2321 - 4759
www.ijmsi.org Volume 2 Issue 4 ǁ April. 2014ǁ PP-04-14
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An Exposition on Solution Process and Nature of Solutions of
Simple Delay Differential Equations
Ukwu Chukwunenye
Department of Mathematics University of Jos P.M.B 2084 Jos, Nigeria
ABSTRACT: This article gave an exposition on the solutions of single-delay autonomous linear differential
equations with algorithmic presentations of the solution processes of two solution methods, with illustrative
examples verifying the consistency of both solution methods. The article went further to examine the nature of
solutions, obtaining in the process a complete characterization of all non-oscillatory solutions of a problem
instance in the absence of initial functions. In the sequel the article proved that there is no nontrivial oscillatory
solution if a constant initial function is specified for the delay differential equation in question and hence for
delay differential equations in general.
KEYWORDS: Algorithmic, Continuity, Delay, Nature, Process.
I. INTRODUCTION
The method of steps and forward continuation recursive procedure are prevalent in the literature on
functional differential equations. However literature on detailed illustrative examples and robust analyses of
problem instances is quite sparse, leaving very little room for enhanced understanding and appreciation of this
class of differential equations. This article, which makes a positive contribution in the above regard, is
motivated by Driver (1977) and leverages on some given problems there to conduct detailed analyses on the
process and structure of solutions and thorough analyses of some problem instances. The algorithmic format
used in the two solution methods, the verification of the consistency of both methods, the investigation of
appropriate feasibility conditions on the solutions and the lucidity of our presentations are quite novel in this
area where presentations are skeletal for the most part, forcing readers to plough through several materials and
figure out a lot of things on their own with the attendant opportunity cost in time. See also Hale (1977) for more
discussion on method of steps.
II. PRELIMINARIES
For an arbitrary initial function problem:
      (1)x t ax t bx t h  
   ( ) , , 0 (2)x t t t h  
scalar constants continuous, , ,a b  it is clear that the integrating factor,   a t
I t e

If we denote the solution on the interval   1 , , 1, 2,....k
J k h kh k   by  y tk , then
on 1
J , (1) becomes      ,x t ax t b t h   from which it is clear the integrating factor,   .a t
I t e

Hence  1
y t can be obtained from the relation:
   1 1
. (3)a t a t
y t e be t h dt C
    
 1
y t exists, since  a t
b e t h
 is integrable, in view of the fact that e a t
and  are continuous, t h
being in  , 0 .h
The solutions  , 1, 2,...,k
y t k  are given recursively by:
   1
(4)a t a t
k k k
y t e b e y t h dt C

    
where the continuity of yk1guarantees the integrability of  be y t ha t
k

 1 and hence the continuity of
yk , assuring the existence of yk , for 1,2,k  

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Above solutions are unique, arising from the unique determination of the skC from the relation:
     1
1 1 (5)k k
y k h y k h
  
Note that if  1, 2,..., 1 ,j k  there is no general expression for the k jc appearing in (6) below. Moreover
if the initial function in (2) is not constant, it is impossible to express yk , in
the form,
1
1
( ) ( ).
k
j at
k k k j
j
y t d d t e t

  
   
   
 However, if a  0 and ( )t is a polynomial of degree m,
say, m  0, then the y sk ' take the form:
   
1
00
6
m k
j j at
k k j k j
jj
y t d t c t e


 
   
 

for some constants 0,1, 0,1, , 1., ; ,k j k jj j kd m c   In other words,  ky t is the sum of some
polynomial of degree m and the product of
at
e and some polynomial  1kP t of degree 1,k 
where    0
y t t .
The process of getting the above coefficients is easy, but the computations get rather unwieldy. The
computational procedure is set out in the following algorithmic steps, noting that:
   k
y t x t for
  1 , . (7)t k h kh 
Method 1: Algorithmic Steps:
[1] Set k = 1 in (6)
[2] Plug in  1
y t for    1
,x t y t for  x t and   t h for  x t h in (1).
[3] Compare the coefficients of the resulting left and right hand expressions to secure the , sk jc and , sk jd
in the relevant ranges for ,j and obtain  1
y t . Verify that the condition    1
0 0y  is satisfied,
where    0
y t t .
[1] Set k = 2 in (6).
[2]          2 2 1
Substitute for , for and for ( ) in (1)y t x t y t x t y t h x t h  
[3] Compare the coefficients of the resulting left and right hand expressions to secure the , sk jc and , sk jd
in the relevant ranges for ,j and hence obtain 2 ( ).y t Check that the condition    2 1
y h y h is
satisfied.
This implementation process continues, so that at the
th
k stage,  k
y t is substituted for
     , fork
x t y t x t  and the already secured  1k
y t h
 substituted for  x t h in (1). Then the
coefficients of the resulting left and right hand expressions are compared to yield the , sk jc and sk jd for
0,1,2,...., , 1,j m m  and hence  , for 3y t k  . Verify that the continuity condition
     1
1 1k k
y k h y k h
   holds.
Alternatively, proceed as follows:
Method 2: Algorithmic Steps:
[1] Set k = 1 in (76) and then    0
y t h t h   and evaluate the integral.
[2] Set    1 0
(0) 0 0y y  to secure 1
,C following the computation of the integral on the right of (4)

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Having initialized the process in step 1,  y tk is obtained by using (4) in conjunction with the
condition      1
1 1 , 2,3,k k
y k h y k h k
    . The integral in (4) can be determined since
 y t hk 1 is already known. The condition      1
1 1k k
y k h y k h
   secures kC .
Example 1
Solve the initial function problem:
      .1 on [0, 2],x t ax t bx t a    R
     1 , 1, 0x t t t t    
Solution:
Method 1
 1 10 1 10 1 1 11 1
1, 1, , ;a t
h k c c d d y t d d t c e       
     1 1 0
y t a y t b y t h  
 
   
11 1 1 11 1
11 1 11 11 1 11 2
1 1
1
0 , .
a t a t
d a c e a d a d t b t a c e
b b
d a d a d b t d d d
a a a
       
 
        
If 1 1 2
0, then 1 1 1.
,
b
a c d a
a
      The condition    1 1 1
0 0 1y d c    .
Verification
1 1 2 2
1 1, 0.
b b
c d a
a a
     
Therefore,  
2
1 2
1 atb b b
y t t e
a a a
    
 
  
 
, if 0;a  that is
  2 2
1 a tb b b
x t t e
a a a
    
 
 
 
for  0 1 ,t  if a  0.
The condition b  0 precludes degeneracy of (1) to an ordinary differential equation.
   2 2 21 2 21
a t
y t d d t c c t e    , where 2 2,0d d
     2 2 1
1y t a y t b y t  
   
21 2 21 21 2 21 2 21
1
1 11 11
at at a t a t a t
a t
d ac e c e ac t e a d a d t ac e ac te
b d d t c e
 
 
       
   
     21 2 1 11 21 11 21 1
0
a a t
d a d b d d b a d bd t c bc e e

        
We claim that the members of the set 1, , at
t e are linearly independent for 0.a 
Proof:
1
0 1 0, for 0.
0 0
at
at at
at
t e
ae ae a
ae
 
 
   
 
 
This proves the claim. Indeed 1, , ,andat at
t e te are linearly
independent as the Wronskian turns out to be
4
0 for 0.a a 
By the above claim it is valid to set each of the above coefficients to zero. Consequently,

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2
11
21 21 12 2
2 2 2 2 2
21 11 1
2 2 2 3 2
, 1 ,
1
2 .
a ab d b b b b
d c c b e b e
a a a a a
d bd bd b b b b b
d
a a a a a a a
    
          
   
  
      
 
The requirement 2 1
(1) (1)y y must be enforced.
2
2 1 2 21 2 21 1 11 1 2 2
(1) (1) ( ) a a a ab b b
y y d d c c e d d c e e e
a a a
            
2 2 2 2
2 2 2 3 2 2 2
2 .a a ab b b b b b b
c e e b e
a a a a a a a
 
          
 
Hence,
2 2 2
( 1) ( 1)
2 3 2 2 2 2
2 2
( 1) ( 1) ( 1) ( 1)
3 2 2
for [1, 2].
( ) ( ) 2
2 1 ,
a t a t at at
a t a t a t a t
t
b b b b b b
x t y t t e e e e
a a a a a a
b b b
e be e b te
a a a
 
   

       
 
     
 
2 2 2 2 2
2 3 2 2 2 2 3 2 2 2 2
(1) 2 2a a a ab b b b b b b b b b b b
y e e b b e e
a a a a a a a a a a a a
                
1 2 2
(1) .a ab b b
y e e
a a a
    Therefore the condition 2 1
(1) (1)y y is satisfied.
We proceed to show that 2 ( )y t satisfies the delay differential equation on [1, 2].
2 2 2
( 1) ( 1) ( 1) ( 1) ( 1) ( 1)
2 2 2
2 2
( 1) ( 1) ( 1)
2
2
( )
.
a t a t at at a t a t a t a t
a t a t a t
b b b b b
y t e be ae e e abe e be
a a a a a
b b
e abte te
a a
     
  
        
  

2 2 2 2
( 1) ( 1) ( 1) ( 1)
2 1 2 2 2
2 2 2 2 2 2
( 1) ( 1) ( 1) ( 1) ( 1)
2 2
( ) ( 1) 2 2
.
a t a t at at a t a t
a t a t a t a t a t
b b b b b b
ay t by t t e be ae e e abe
a a a a a a
b b b b b b
e abte te t be e
a a a a a a
   
    
          
 
         
 
Therefore 2 2 1( ) ( ) ( 1).y t ay t by t   This completes the proof that 2 ( )y t solves the initial function
problem on the interval [1, 2].
Method 2: Algorithmic Steps
Set k = 1 in (4) and then    0
y t h t h   , evaluate the integral.
Set    1 0
(0) 0 0y y  to secure 1
,C following the computation of the integral on the right of (4)
Having initialized the process in step 1,  y tk is obtained by using (4) in conjunction with the
condition      1
1 1 , 2,3,k k
y k h y k h k
    . The integral in (4) can be determined since
 1k
y t h
 is already known. The condition      1
1 1k k
y k h y k h
   secures Ck .
Consider the interval 1 [0, 1].J  Set 0 0
( 1) ( 1) 1 1 , on [ 1, 0].y t t t t J         Apply (4) with
1, 1k h  to get  1 1 12
.a t a t at at atb b
y t e b e t dt C e t e e C
a a
  
     
      


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   1 1 1 12 2 2 2
0 (0) 1 1 1 1 atb b b b b
y C C y t t e
a a a a a
              
 
 
 
 1 2 2
1
1 .atb b
y t t e
a a a
     
 
 
 
Consider the interval 2 2 1 2 2
Set[1, 2]. (1) (1) .a ab b b
J y y e e
a a a
       Apply (4), with 2,k  to get
    ( 1)
2 1 2 22 2
1 ( 1) 1a t a t a t a t a tb b b
y t e b e y t dt C e b e t e dt C
a a a
  
         
             
 
 
2 2 2 2
2 22 2
a t at at at a ab b b b
y t e e dt e tdt e dt e b dt e dt C
a a a a
    
       
 
 
 
    
 
2 2 2 2
2 23 2 2 2
1 1at at at a a at atb b b b
y t e e e e bt e t te e C
a a a a a a
     
        
  
  
  
2 2 2 2
2 22 2 3 2 2 2
1 1
(1) a a a a a a a a ab b b b b b b
y e e e e e e b e e e C
a a a a a a a a a
     
            
  
  
  
2 2 2 2 2
22 2 3 2 2 2 3
2 2
22 2 3 2
2
a a a
a a a
b b b b b b b b
e e b C e
a a a a a a a a
b b b b b
e e b C e
a a a a a
           
        
2 2
2 2 2 3 2
1 2a a a a ab b b b b
C e e e be e
a a a a a
    
        
 
2 2 2 2
2 3 2 2 2
1 1at at at a a at atb b b b
y t e e e e bt e t te e
a a a a a a
     
       
  
  
  
2 2
2 2 3 2
1 2at a a a a ab b b b b
e e e e be e
a a a a a
    
       
 
 
 
 
2 2 2 2 2
( 1) ( 1)
2 3 2 2 2 3
a t a tb b b b b
y t e bt te t
a a a a a
 
      
2 2
2 2 3 2
1 2at a a a a ab b b b b
e e e e be e
a a a a a
     
        
 
 
2 2 2 2 2
( 1) ( 1)
2 3 2 2 2 3
2 2
( 1) ( 1) ( 1) ( 1) ( 1)
2 2 3 2
2
a t a t
a t a t at at a t a t a t
b b b b b
y t e bt te t
a a a a a
b b b b b
e e e e e be e
a a a a a
 
    
      
      
 
2 2 2 2 2
( 1) ( 1)
2 3 2 2 2 3
2 2
( 1) ( 1) ( 1) ( 1) ( 1)
2 2 3 2
2 2 2 2 2 2
( 1) ( 1)
3 2 2 2 2 2 3 2
2
2 1 2
a t a t
a t a t at at a t a t a t
at a t a t
b b b b b
y t e bt te t
a a a a a
b b b b b
e e e e e be e
a a a a a
b b b b b b b b b
t e b e b te
a a a a a a a a a
 
    
 
      
      
    
               
     

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2 2 2 2 2 2
( 1) ( 1)
3 2 2 2 2 2 3 2
( ) 2 1 2at a t a tb b b b b b b b b
x t t e b e b te
a a a a a a a a a
     
                
     
is the solution of the initial function problem on the interval [1, 2]. This is consistent with the result obtained
using Method 1.
The case 0a  ( ) ( 1), 0; ( ) ( ) 1 , [ 1,0].x t bx t t x t t t t         Hence
2
1
1
( ) ( ) , [0,1].
2
x t bt x t bt d t     The continuity condition (0) (0) 1x  
2
1 1
1
1 ( ) ( ) 1 on [0,1].
2
d x t y t bt     
On (1,2),
2 2 2 3
2 2
1 1
( ) ( 1) ( ) ( ) ( 1) .
2 6
x t b t b x t y t b t bt d        
The continuity condition
2 3
2 1 2 2
1 1
(1) (1) 1 ( ) ( ) ( 1) 1.
2 6 2
b
y y d b x t y t b t bt          
Using the relation    1
with 0, 1 and 1 yields,a t a t
k k k
y t e b e y t h dt C a h k

       
       
 
1 1
1
2
0 1 1 1 1 1
2
01 ; (0) 1 1
2
( ) 1, [0,1].
2
y t y
y t
t
by t h dt C b t dt C btdt C b C C
t
x t b t
              
    
  
   
     
2
2 1 2
2 2 3
1 2 2 2
2 3
2 2
1 1
1 1
( 1) 1 ( 1) ;
2 6
1 1 1 1
1 1 ( ) ( 1) 1 , [1,2].
2 2 6 2
y t
y y y t
by t h dt C b b t dt C b t bt C
b C b C b x t b t bt b t
          
               
 
  
 
This also agrees with the preceding result.
Nature of nontrivial oscillatory solutions
One can appreciate the tedium involved in extending the solutions beyond the interval [1, 2]. Equation
(1) is a first order scalar, linear, homogeneous delay differential equation with real coefficients. Solutions
obtained thus far have been real non-oscillatory.
In general oscillatory solutions of (2) can be obtained in the absence of initial function specification by
assuming solutions of the form:
  , (8)t
x t e

to obtain the transcendental equation:
0 (9)h
a de 
 
 
This transcendental equation is analyzed completely to determine the nature of nontrivial solutions of oscillatory
type.
We assert that in general real solutions do not exist in the form   t
x t e
 .
Our analysis and conclusion in the next example will validate the above assertion.
Example 2
Prove that:  ( ) 2 (10)x t x t   
admits non-trivial oscillating solutions and determine the nature of such solutions.

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Prove also that there are no nontrivial oscillatory solutions if a constant initial function is specified for the delay
differential equation (10).
Proof
Our proof will rely partly on the following result which we proceed to establish:
 ln 1 1 (11)x x x   
To establish this result, observe that the result is equivalent to:
1 (12)x
e x 
Set:
   1 (13)x
f x e x  
 Then,
1 0, 0
0, 0
x
e x
f x
x
    
  
  
 0 0, ( ) 0 iff 0f f x x   . Thus (0, 0) is the only stationary point of f(x).
Now   f x e xx
( ) , . In particular  0 0f    0,0 is a global minimum of  f x .
Hence   0 ;f x x  that is 1 ,x
e x x   ; this translates to  ln 1x x  .
Assume a solution of the form x t e t
( )  
in the example. Then 2
e



  , noting that
a b  0 1, and
2
h

 .
The equation:
2
(14)e



 
is not satisfied for any real 0  , since
2
0e

  if  is real. Therefore if  is real, then   0, so
that   
 
e /2
 ln / 2    
But  ln 1 0        , by an appeal to (11).
Therefore,     1 2/ , or 1 1
2

 
 
 
 
. Since 2 1 0,   we conclude that
2
0
2


 

This contradicts the fact that  must be negative. Therefore, the equation  



e 2 cannot be satisfied
by any real  . In other words, any solution of the form   t
x t e
 must yield i    , when  and 
are real numbers such that   0. Plugging this  into the transcendental equation yields:
2
cos sin , (15)
2 2
i e i

 
 
     
       
    
from which we infer that:
2
cos (16)
2
e



  
   
 

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and:
 2
sin 17
2
e



  
   
 
If   0, the above relations imply that:
cos 0 (18)
2
 
 
 
and:
sin (19)
2


 
   
 
From (18) we infer that:  2 1 ,
2 2
k
 
  k integer
2 1 (20)k  
  is an odd integer, positive or negative.
 19  the relation 1  is infeasible.
Therefore: 1 (21) 
The relations (20) and (21) imply that   1, thus    1 or 1; hence if   1, then    1or1 and
  1 2cos sin , (22)x t c t c t 
for real constants c1 and c2.
(22) can be rewritten in the form:
   2 2
1 2 sin (23)x t c c t   
for some real δ with 0
2
  . Clearly if c1 and c2 do not vanish at the same time, then
2 2
1 2
0c c  , so
that (23) represents non-trivial oscillatory solutions, being sinusoidal.
Remarks
If   2
0, 8 e

 

      0 since 2
0e


  .
If  0, then 0  , since real solutions do not exist in the form (8)
If 0  , then cos 0
2


 
 
 
   1 4 3 4
2 2 2
k k
  
    
1 4 3 4k k     , for any integer k. If  0, then cos 0
2

 
 
 
 
4 1 4k k    (1st
quadrant) or 3 4 4 4k k    (4th
quadrant for any integer k).
(16) and (17) imply that 2
cos
2
e


 
 
  
 
and 2
sin
2
e


 
 
  
 
      2 2
1e e       2 2
1e       
      2 2 2 2
e e

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   
  
e 2
 solutions to the transcendental equation (9) are given by     i and the set
       
, : ,   
e 2
0 . There is an infinity of oscillatory solutions. These solutions are
given by:
        1 2
24cos sin ,t
x t e c t c t
  
for all real  , for which   2 2
 
e and
2
.e  

 These solutions are nontrivial if
c c1
2
2
2
0  . Furthermore, if   0 then   0 and  tan   
2   . If   0, then
1 4 3 4k k    and if   0, then, 4 1 4k k   , for any integer k. If  0, then 0  
and  
 2 2
2 2 2
3
1 .
2 ! 2
j
j j
e    
    


 
       Therefore
2
0  if 0. 
If 0,  the relation   2 2
 
e is infeasible for
2 2
1, since 1e 
   
    . In fact it is
feasible for 0 0.45445  . In other words,
2
0  , for 0 0.45445  . Clearly e
 
1 for
  0. In particular 0 12
  for 0 045445  . . Hence,     0 1 1, for this set of  -
values.
From the preceding analysis, we have obtained a complete characterization of all non-oscillatory solutions of
(10) in the absence of initial functions.
Next we will prove that there is no nontrivial oscillatory solution if a constant initial function is specified for
the delay differential equation (10).
Suppose  is a continuous initial function specified by      x t t t   , ,2 0 with respect to (10).
Then (24)    1
0 0x c    . If   0, so that 1,   then
 x t c t c t 1 2cos sin    2
2 2
x c       and      1 2
2
.
4 42
x c c     
Suppose   0,t  a constant. Then,  c c c c1 2 1 2
2
2
   
 1 1 1 1 1 2
2
2 0 0
2
c c c c c c             x t t0 2
  trivial solution.
Suppose   0, so that 4 1 4k k   or 3 4 4 4   k k , k integer.
Recall that  1 2 0( ) cos( ) sin( ) , , ; ( ) , , 0 .
2 2
t
x t e c t c t t x t t  
  
    
           
Let us evaluate ( )x t for all t such that    ,
4 2
,
4 2
.t t
 
 
 
       Since
, ,
2
t
 
   
these requirements are simultaneously satisfied only by 1.  Therefore,
for feasibility take k  0 such that , 0
2 2
 

  
   
and
4
, 0
2



 
 
  
. From the preceding results,

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 x e c





   







2
02
2 0 and  1 2
4
0
4
2
2
c cx e








 
  
 
But   4 4 4 4
1 0 2 2 2 2
2 2 2
0 1 0,
2 2 2
x c c e c e c e e c
   
   

 
  
              
2
4 4 4 42 2
0 or 1 0. But 1 0
2 2
c e e e e
   
   
 
       
      
                  
if and only if
4 4 2
.
2
e e
 
 

 
Let us examine the function
0, if 0
2
. Clearly, 0, if 0
2
0, if 0
v v v v
v
dw
w e e e e v
dv
v
 
 

      
 
Therefore w is decreasing for v < 0 and increasing for v > 0; w has global minimum at v = 0, with
minimum value 2 2 0.585786.  (Note: ( ) 0w v v   R ). We deduce that w is never less that
2 2. This proves that 4 42
1 0,
2
e e
 
 

  
  
      
proving that 2 0c  ; hence 1 0.c 
We conclude that ( ) 0, ,
2
x t t

   which implies that only the trivial solution exists.
For the case 0,  we earlier proved that and for feasibility.0 0.45445 1 1,     
We wish to evaluate ( )x t for all t such that    , , .
4 2 4 2
t t
   

 
       Since
, ,
2
t

  
 
 
these requirements are simultaneously satisfied only by 1.  Since 1 1,   the case
0  is infeasible. These prove that there are no nontrivial oscillatory solutions if a constant initial function is
specified for the delay differential equation (10).
Remarks:
One could also reason along the same lines as in Driver (1977) to establish the existence of infinitely many
complex roots for the equation 2 .e




  To achieve this, set
1
, ( ) 0, where ( ) 1 ,
h
z
z w z w z ze


    and h is an arbitrary delay. Clearly ( ) 1,for 0.w z z  But
( )w z has an isolated singularity at 0.z  Thus by Picard’s theorem, in every neighbourhood of
0, ( )z w z takes on every value infinitely often with the exception of 1. Hence ( ) 0w z  infinitely often and so
the equation 2
e




  must have infinitely many complex roots: ; ,i      real. However the
subsequent analyses undertaken above are imperative for solving the problem.
III. CONCLUSION
This article gave an exposition on how single-delay scalar differential equations could be solved using
two methods presented in algorithmic formats. The results, which relied on the continuity of solutions, verified
the consistency of both methods using the same problem, as well as gave a clue on the associated level of
difficulty. It went on to conduct detailed and rigorous analyses on the nature, characterization and flavor of
solutions for some problem instances. The article revealed that functional differential equations are difficult to

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solve, even for the most innocent of problems-such as the autonomous linear delay type with further simplifying
restrictions; this difficulty is largely attributable to the many issues that must be dealt with, two such issues
being the transcendental nature of the associated characteristic equations and the prescribed class of initial
functions.
REFERENCE
[1] Driver, R. D. (1977). Ordinary and Delay Differential Equations. Applied Mathematical Sciences 20,
Springer-Verlag, New York.
[2] Hale, J. K. (1977). Theory of functional differential equations. Applied Mathematical Science, Vol. 3,
Springer-Verlag, New York.